This is question 4 of Exercise 8(c) in A Second Course in Mathematical Analysis (J. C. Burkill, H. Burkill).
Let $I$ be the interval $[-1,2] \times [-1,0] \times [0,2]$ in $\mathbb{R}^3$. Show that $\displaystyle\iiint_I \left(\dfrac{z}{1-|x|y}\right)^2 dzdydx = 9 \log 6.$
I have attempted the question but not obtained the given answer. I can write the integral as iterated integrals $$ I = \int_{-1}^2 \int_{-1}^0 \int_{0}^2 \dfrac{z^2}{(1-|x|y)^2} dz dy dx $$
I can evaluate $\int_0^2 z^2 dz = 8/3$. I include a WolframAlpha link for verification.
This means that I can proceed by writing $$ I = \int_{-1}^2 \int_{-1}^0 \dfrac{8}{3} (1-|x|y)^{-2} dy dx.$$
Now, I can evaluate $\int_{-1}^0 (1-|x|y)^{-2} dy = (1 + |x|)^{-1}.$
Finally, the last integral is $\displaystyle\int_{-1}^2 \dfrac{8}{3}\big(1+|x|\big)^{-1} dx$ which I do by splitting it into two integrals, one from $-1$ to $0$ and the other from $0$ to $2$. This gives me $\dfrac{8}{3} \log 6$.
I could check my final result using WolframAlpha, which seems to agree.
Could someone please let me know if this is a misprint in the book or if there is a flaw in my working? Thank you!